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Recall that #\sqrt{\blue {9}} = \orange 3#.
To find the square root, we are looking for #\orange{\text{a non-negative number}}#, which will give us #\blue{\text{the number under the root sign}}# once squared.
The (square) root #\sqrt{\blue a}# of #\blue a# is the #\orange x# for which we have
\[\orange x^2=\blue a \qquad \text{and} \qquad \orange x\geq 0\]
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Example
\[\begin{array}{c}
\sqrt{\blue {9}}\;=\;\orange3 \\ \text{ because } \\ \orange 3^2=\blue {9} \quad \text{and}\quad \orange 3 \geq 0\\ \\ \sqrt{\blue {0}}\;=\;\orange0 \\ \text{ because } \\ \orange 0^2=\blue {0} \quad \text{and}\quad \orange 0 \geq 0\end{array}\]
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The square root of a number cannot be negative.
Hence, #\sqrt{\blue 9}=\orange 3# and not #\orange{-3}#, although we have #(\orange{-3})^2=\blue 9#.
It is defined like this to define the square root unambiguously.
Hence, we can speak of the square root of #\blue {9}#.
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\[\begin{array}{c}
\sqrt{\blue {9}} \;\red{\neq}\; \orange{ -3} \\ \text{ because } \\
\orange {-3} {\;\red{\not\geq}\;} 0
\end{array}\]
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The root of a non-negative number #\blue {a}# is a solution of the equation #\orange {x}^2=\blue {a}# in the unknown #\orange {x}#. This means that #\orange {x}=\sqrt{\blue {a}}# satisfies this equation, so #\left(\sqrt{\blue {a}}\right)^2 = \blue {a}#.
If #\blue {a}\gt0#, then there is a second solution of the equation #\orange {x}^2=\blue {a}#, namely #\orange {x}=-\sqrt{\blue {a}}#. After all, #\left(-\sqrt{\blue {a}}\right)^2 = \blue {a}#. This solution is negative.
A solution of an equation is called a root of the equation. So there may be two roots of the equation #\orange {x}^2=\blue {a}#.
To ensure that #\sqrt{\blue {a}}# is uniquely determined, we demand that this expression is not negative. Thus, #\sqrt{\blue {a}}# is the only non-negative root of the equation #\orange {x}^2=\blue {a}# and #-\sqrt{\blue {a}}# is the only negative root of that equation if #\blue {a}\ne0#.
The root of a number #\sqrt{\blue {a}}# has a meaning only when #\blue{a}\geq 0#. There is no real number #x# such that #\orange {x}^2# is negative, because \[\begin{array}{c}
\text{negative} \cdot \text{negative}= \text{positive} \\
\text{positive} \cdot \text{positive} = \text{positive}
\end{array}\]
Therefore, expressions and algebraic rules involving roots are only valid if the expression under the root sign is not negative. For example, for the expression #\sqrt{\green{x y}}#, we should have that #\green{x y}\geq 0#.
Subsitute #x=5# in #\sqrt{x-4}#. Is that possible?
yes
We can substitute every value for #x# for which #x-4# is greater than or equal to #0# since the number below the root sign cannot be negative.
If we substitute #x=5#, we get #5-4=1 \geq 0 #. Hence, we can substitute this value.
Square roots
